Problem: $ B = \left[\begin{array}{r}-2 \\ 5 \\ 0\end{array}\right]$ $ E = \left[\begin{array}{rr}3 & 3\end{array}\right]$ What is $ B E$ ?
Explanation: Because $ B$ has dimensions $(3\times1)$ and $ E$ has dimensions $(1\times2)$ , the answer matrix will have dimensions $(3\times2)$ $ B E = \left[\begin{array}{r}{-2} \\ {5} \\ \color{gray}{0}\end{array}\right] \left[\begin{array}{rr}{3} & \color{#DF0030}{3}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ B$ , with the corresponding elements in column $j$ of the second matrix, $ E$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ B$ with the first element in ${\text{column }1}$ of $ E$ , then multiply the second element in ${\text{row }1}$ of $ B$ with the second element in ${\text{column }1}$ of $ E$ , and so on. Add the products together. $ \left[\begin{array}{rr}{-2}\cdot{3} & ? \\ ? & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ B$ with the corresponding elements in ${\text{column }1}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{-2}\cdot{3} & ? \\ {5}\cdot{3} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ B$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ E$ and add the products together. $ \left[\begin{array}{rr}{-2}\cdot{3} & {-2}\cdot\color{#DF0030}{3} \\ {5}\cdot{3} & ? \\ ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{-2}\cdot{3} & {-2}\cdot\color{#DF0030}{3} \\ {5}\cdot{3} & {5}\cdot\color{#DF0030}{3} \\ \color{gray}{0}\cdot{3} & \color{gray}{0}\cdot\color{#DF0030}{3}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-6 & -6 \\ 15 & 15 \\ 0 & 0\end{array}\right] $